And 4 72 12 172 optimal y1 92 y2 0 y3 52 apply

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Unformatted text preview: 1 -2 1 -2 -2 -7/2 -1/2 -9/2 0 1 0 0 0 1 0 0 0 0 1 0 -1 -3/2 -1/2 -5/2 3 2 -1 6 4 7/2 1/2 17/2 ← pivot row 1 -2 0 -2 1 -2 0 -2 1 0 0 0 1 0 0 0 -2 -3 -2 -5 0 0 1 0 -1 -2 1 -2 -2 -7/2 -1/2 -9/2 0 1 0 0 0 1 0 0 0 0 1 0 -1 -3/2 -1/2 -5/2 3 2 -1 6 4 7/2 1/2 17/2 ← pivot row optimal 1 -2 0 -2 1 0 0 0 0 0 1 0 -2 -7/2 -1/2 -9/2 0 1 0 0 -1 -3/2 -1/2 -5/2 4 7/2 1/2 17/2 optimal 1 -2 0 -2 1 0 0 0 0 0 1 0 -2 -7/2 -1/2 -9/2 x1 0 x2 = 4 x3 1/2 0 1 0 0 -1 -3/2 -1/2 -5/2 4 7/2 1/2 17/2 optimal 1 -2 0 -2 1 0 0 0 0 0 1 0 -2 -7/2 -1/2 -9/2 x1 0 x2 = 4 x3 1/2 0 1 0 0 -1 -3/2 -1/2 -5/2 and 4 7/2 1/2 17/2 optimal y1 9/2 y2 = 0 , y3 5/2 1 -2 0 -2 1 0 0 0 0 0 1 0 -2 -7/2 -1/2 -9/2 x1 0 x2 = 4 x3 1/2 0 1 0 0 -1 -3/2 -1/2 -5/2 Optimal value = −17/2. and 4 7/2 1/2 17/2 optimal y1 9/2 y2 = 0 , y3 5/2 Apply the dual simplex algorithm to the following problem. P maximize −4x1 − 2x2 − x3 subject to − x1 − x2 + 2x3 ≤ −3 −4x1 − 2x2 + x3 ≤ −4 x1 + x2 − x3 ≤ 2 0 ≤ x1 , x2 , x3 . -1...
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This note was uploaded on 01/29/2014 for the course MATH 407 taught by Professor Staff during the Fall '08 term at University of Washington.

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